How to sort an array of integers correctly

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Trying to get the highest and lowest value from an array that I know will contain only integers seems to be harder than I thought.

var numArray = [140000, 104, 99];
numArray = numArray.sort();
console.log(numArray)

I'd expect this to show 99, 104, 140000. Instead it shows 104, 140000, 99. So it seems the sort is handling the values as strings.

Is there a way to get the sort function to actually sort on integer value?

32 Answers

By default, the sort method sorts elements alphabetically. To sort numerically just add a new method which handles numeric sorts (sortNumber, shown below) -

var numArray = [140000, 104, 99];
numArray.sort(function(a, b) {
  return a - b;
});

console.log(numArray);

Documentation:

Mozilla Array.prototype.sort() recommends this compare function for arrays that don't contain Infinity or NaN. (Because Infinity - Infinity is NaN, not 0).

Also examples of sorting objects by key.

I am surprised why everyone recommends to pass a comparator function to sort(), that makes sorting really slow!

To sort numbers, just create any TypedArray:

var numArray = new Float64Array([140000, 104, 99]);
numArray = numArray.sort();
console.log(numArray)

array.sort does a lexicographic sort by default, for a numeric sort, provide your own function. Here's a simple example:

function compareNumbers(a, b)
{
    return a - b;
}

numArray.sort(compareNumbers);

Also note that sort works "in place", there's no need for the assignment.

The reason why the sort function behaves so weird

From the documentation:

[...] the array is sorted according to each character's Unicode code point value, according to the string conversion of each element.

If you print the unicode point values of the array then it will get clear.

console.log("140000".charCodeAt(0));
console.log("104".charCodeAt(0));
console.log("99".charCodeAt(0));

//Note that we only look at the first index of the number "charCodeAt(  0  )"

This returns: "49, 49, 57".

49 (unicode value of first number at 140000)
49 (unicode value of first number at 104)
57 (unicode value of first number at 99)

Now, because 140000 and 104 returned the same values (49) it cuts the first index and checks again:

console.log("40000".charCodeAt(0));
console.log("04".charCodeAt(0));

//Note that we only look at the first index of the number "charCodeAt(  0  )"

52 (unicode value of first number at 40000)
40 (unicode value of first number at 04)

If we sort this, then we will get:

40 (unicode value of first number at 04)
52 (unicode value of first number at 40000)

so 104 comes before 140000.

So the final result will be:

var numArray = [140000, 104, 99];
numArray = numArray.sort();
console.log(numArray)

104, 140000, 99

Conclusion:

sort() does sorting by only looking at the first index of the numbers. sort() does not care if a whole number is bigger than another, it compares the value of the unicode of the digits, and if there are two equal unicode values, then it checks if there is a next digit and compares it as well.

To sort correctly, you have to pass a compare function to sort() like explained here.

just do .sort((a, b) => a - b) instead of .sort() itself

var numArray = [140000, 104, 99];
numArray.sort((a, b) => a - b);
console.log(numArray)

Ascending

arr.sort((a, b) => a - b);

Descending

arr.sort((a, b) => b - a);

Just for fun:

Descending = Ascending + Reverse

arr.sort((a, b) => a - b).reverse();

In JavaScript the sort() method's default behaviour is to sort values in an array alphabetically.

To sort by number you have to define a numeric sort function (which is very easy):

...
function sortNumber(a, b)
{
  return a - b;
}

numArray = numArray.sort(sortNumber);

The function 'numerically' below serves the purpose of sorting array of numbers numerically in many cases when provided as a callback function:

function numerically(a, b){
    return a-b;
}

array.sort(numerically); 

But in some rare instances, where array contains very large and negative numbers, an overflow error can occur as the result of a-b gets smaller than the smallest number that JavaScript can cope with.

So a better way of writing numerically function is as follows:

function numerically(a, b){
   if(a < b){
      return -1;
   } else if(a > b){
      return 1;
   } else {
      return 0;
   }
}

to handle undefined, null, and NaN: Null behaves like 0, NaN and undefined goes to end.

array = [3, 5, -1, 1, NaN, 6, undefined, 2, null]
array.sort((a,b) => isNaN(a) || a-b)
// [-1, null, 1, 2, 3, 5, 6, NaN, undefined]

Array.sort uses alphabetic sorting instead of numeric.

To support numbers, add like following

var numArray = [140000, 104, 99];
numArray.sort((a, b) =>  a - b); // <-- Ascending
numArray.sort((a, b) =>  b - a); // <-- Descending
console.log(numArray);

OUTPUT :

Array Numeric Sorting

While not required in JavaScript, if you would like the sort() compareFunction to strictly return -1, 0, or 1 (similar to how the spaceship operator works in PHP), then you can use Math.sign().

The compareFunction below strictly returns -1, 0, or 1:

numArray.sort((a, b) => Math.sign(a - b));

Note: Math.sign() is not supported in Internet Explorer.

For a normal array of elements values only:

function sortArrayOfElements(arrayToSort) {
    function compareElements(a, b) {
        if (a < b)
            return -1;
        if (a > b)
            return 1;
        return 0;
    }

    return arrayToSort.sort(compareElements);
}

e.g. 1:
var array1 = [1,2,545,676,64,2,24]
**output : [1, 2, 2, 24, 64, 545, 676]**

var array2 = ["v","a",545,676,64,2,"24"]
**output: ["a", "v", 2, "24", 64, 545, 676]**

For an array of objects:

function sortArrayOfObjects(arrayToSort, key) {
    function compareObjects(a, b) {
        if (a[key] < b[key])
            return -1;
        if (a[key] > b[key])
            return 1;
        return 0;
    }

    return arrayToSort.sort(compareObjects);
}

e.g. 1: var array1= [{"name": "User4", "value": 4},{"name": "User3", "value": 3},{"name": "User2", "value": 2}]

**output : [{"name": "User2", "value": 2},{"name": "User3", "value": 3},{"name": "User4", "value": 4}]**

The accepted answer and equivalents like numArray.sort((a,b) => a - b) are great when the array contains only numbers without infinities or NaN. They can be extended to handle infinities and NaN like so:

numArray.sort((a,b) => (+a || 0) - (+b || 0) || 0);

This sorts NaN (or any non-number, like 'foo' or {}) as if it were 0. The final || 0 is needed to handle the case where a and b are equal infinities.

Try this code as below

var a = [5, 17, 29, 48, 64, 21];
function sortA(arr) {
return arr.sort(function(a, b) {
return a - b;
})
;} 
alert(sortA(a));

TypeScript variant

const compareNumbers = (a: number, b: number): number => a - b

myArray.sort(compareNumbers)

In order to create this kind of sort, you have to pass a function that will check which comes first.

define inside the function which value do you wanna check: a.id - a.id

        const myJson = [
            { id: 1, name: 'one'},
            { id: 4, name: 'four'},
            { id: 2, name: 'two'},
            { id: 3, name: 'three'}
        ];

        // provide the sort method to check
        const myNewSort = myJson.sort(function(a, b) {
          return a.id - b.id;
        });

        console.log('my new sort',myNewSort)

You can sort number array simply by

const num=[13,17,14,19,16];
let temp;
for(let i=0;i<num.length;i++){
    for(let j=i+1;j<num.length;j++){
        if(num[i]>num[j]){
            temp=num[i]
            num[i]=num[j]
            num[j]=temp
        }
    }
}

console.log(num);

Sort integers > 0, think outside the box:

function sortArray(arr) {
  return new Promise((resolve) => {
    const result = []
    arr.forEach((item) => {
      setTimeout(() => {
        result.push(item)
        if (result.length === arr.length) resolve(result)
      }, item)
    })
  })
}

sortArray([4, 2, 42, 128, 56, 2]).then((result) => {
  document.write(JSON.stringify(result))
})

Note that this should not be used productively, .sort() is better suited for this, check the other answers

sort_mixed

Object.defineProperty(Array.prototype,"sort_mixed",{
    value: function () { // do not use arrow function
        var N = [], L = [];
        this.forEach(e => {
            Number.isFinite(e) ? N.push(e) : L.push(e);
        });
        N.sort((a, b) => a - b);
        L.sort();
        [...N, ...L].forEach((v, i) => this[i] = v);
        return this;
    })

try a =[1,'u',"V",10,4,"c","A"].sort_mixed(); console.log(a)

If anyone doesn't understand how Array.sort() works with integers, read this answer.

Alphabetical order:

By default, the sort() method sorts the values as strings in alphabetical and ascending order.

const myArray = [104, 140000, 99];
myArray.sort();
console.log(myArray); // output is [104, 140000, 99]

Ascending order with array.sort(compareFunction):

const myArray = [104, 140000, 99];
myArray.sort(function(a, b){
  return a - b;
});
console.log(myArray); // output is [99, 104, 140000]

Explanation from w3schools:

compareFunction defines an alternative sort order. The function should return a negative, zero, or positive value, depending on the arguments, like: function(a, b){return a-b} When the sort() method compares two values, it sends the values to the compare function, and sorts the values according to the returned (negative, zero, positive) value.

Example:

When comparing 40 and 100, the sort() method calls the compare function(40,100).

The function calculates 40-100, and returns -60 (a negative value).

The sort function will sort 40 as a value lower than 100.

Descending order with array.sort(compareFunction):

const myArray = [104, 140000, 99];
myArray.sort(function(a, b){
  return b - a;
});
console.log(myArray); // output is [140000, 104, 99]

This time we calculated with b - a(i.e., 100-40) which returns a positive value.

You can get height and lowest number simply by using max() and min() in-built function

var numArray = [140000, 104, 99];
console.log(Math.max(...numArray));
console.log(Math.min(...numArray));

If you want to sort in ascending or descending order

numArray.sort((a, b)=> a - b);

Know more

If you need to calculate and sort the largest charCodeAt from a list of string this is the right way.

const arrayLines = '1.1.1.1\n1.0.1.1\n1.1.1.2\n1.1.1.0'.split('\n');

// Response: (4) ['1.0.1.1', '1.1.1.0', '1.1.1.1', '1.1.1.2']
arrayLines.sort((a, b) => {
    let a_charCodeSize = 0,
        b_charCodeSize = 0;

    // Loop true a & b characters and calculate the charCodeAt size.
    for (const aChar of a) a_charCodeSize += aChar.charCodeAt(0);
    for (const bChar of b) b_charCodeSize += bChar.charCodeAt(0);

    return a_charCodeSize - b_charCodeSize;
});

let grade =[80,100,50,90,40];
grade.sort((x,y)=> x-y);
grade.forEach(element=>console.log(element));

As sort method converts Array elements into string. So, below way also works fine with decimal numbers with array elements.

let productPrices = [10.33, 2.55, 1.06, 5.77];
console.log(productPrices.sort((a,b)=>a-b));

And gives you the expected result.

  1. Ascending

    const movements = [200, 450, -400, 3000, -650, -130, 70, 1300];

If we return something < 0 then A will be before B If we return something > 0 then B will be before A

 movements.sort((a, b) => {
      if (a > b) return 1; //- (Switch order)
      if (a < b) return -1; //- (Keep order)
    });

a - current value, b - the next value.

  1. Descending

    movements.sort((a, b) => { if (a > b) return -1; // - (Keep) if (a < b) return 1; // - (Switch) });

! Improve, best solution !

movements.sort ((a, b) => a - b); // Same result!

If a < b it's negative number(Switch) If a < b it's negative number(Keep)

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